Methods for Constructing a Yield Curve

[Pages:12]Methods for Constructing a Yield Curve

Patrick S. Hagan Chief Investment Office, JP Morgan 100 Wood Street London, EC2V 7AN, England, e-mail: patrick.s.hagan@ Graeme West Financial Modelling Agency, 19 First Ave East, Parktown North, 2193, South Africa e-mail: graeme@finmod.co.za, finmod.co.za

Abstract

In this paper we survey a wide selection of the interpolation algorithms that are in use in financial markets for construction of curves such as forward curves, basis curves, and most importantly, yield curves. In the case of yield curves we also review the issue of bootstrapping and discuss how the interpolation algorithm should be intimately connected to the bootstrap itself. As we will see, many methods commonly in use suffer from problems: they posit unreasonable expections, or are not even necessarily arbitrage free. Moreover, many methods result in material variation in large sections of the curve when only one

input is perturbed (the method is not local). In Hagan and West [2006] we introduced two new interpolation methods--the monotone convex method and the minimal method. In this paper we will review the monotone convex method and highlight why this method has a very high pedigree in terms of the construction quality criteria that one should be interested in.

Keywords

yield curve, interpolation, fixed income, discount factors

1 Basic Yield Curve Mathematics

Much of what is said here is a reprise of the excellent introduction in [Rebonato, 1998, ?1.2].

The term structure of interest rates is defined as the relationship between the yield-to-maturity on a zero coupon bond and the bond's maturity. If we are going to price derivatives which have been modelled in continuous-time off of the curve, it makes sense to commit ourselves to using continuously-compounded rates from the outset.

Now is denoted time 0. The price of an instrument which pays 1 unit of currency at time t--such an instrument is called a discount or zero coupon bond--is denoted Z(0, t). The inverse of this amount could be denoted C(0, t) and called the capitalisation factor: it is the redemption amount earned at time t from an investment at time 0 of 1 unit of currency in said zero coupon bonds. The first and most obvious fact is that Z(0, t) is decreasing in t (equivalently, C(0, t) is increasing). Suppose Z(0,t1) < Z(0,t2) for some t1 < t2. Then the arbitrageur will buy a zero coupon bond for time t1, and sell one for time t2, for an immediate income of Z(0,t2)--Z(0,t1) > 0. At time t1 they will receive 1 unit of currency from the bond they have bought, which they could keep under their bed for all we care until time t2, when they deliver 1 in the bond they have sold.

What we have said so far assumes that such bonds do trade, with sufficient liquidity, and as a continuum i.e. a zero coupon bond exists for

every redemption date t. In fact, such bonds rarely trade in the market. Rather what we need to do is impute such a continuum via a process known as bootstrapping.

It is more common for the market practitioner to think and work in terms of continuously compounded rates. The time 0 continuously compounded risk free rate for maturity t, denoted r(t), is given by the relationship

C(0, t) = exp(r(t)t)

(1)

Z(0, t) = exp(-r(t)t)

(2)

r(t) = - 1 ln Z(0, t)

(3)

t

In so-called normal markets, yield curves are upwardly sloping, with longer term interest rates being higher than short term. A yield curve which is downward sloping is called inverted. A yield curve with one or more turning points is called mixed. It is often stated that such mixed yield curves are signs of market illiquidity or instability. This is not the case. Supply and demand for the instruments that are used to bootstrap the curve may simply imply such shapes. One can, in a stable market with reasonable liquidity, observe a consistent mixed shape over long periods of time.

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Z (0, t2)

0 Z (0, t1)

1

t1

t2

1

Figure 1: The arbitrage argument that shows that Z(0, t) must be decreasing.

The shape of the graph for Z(0, t) does not reflect the shape of the yield curve in any obvious way. As already mentioned, the discount factor curve must be monotonically decreasing whether the yield curve is normal, mixed or inverted. Nevertheless, many bootstrapping and interpolation algorithms for constructing yield curves miss this absolutely fundamental point.

Interestingly, there will be at least one class of yield curve where the above argument for a decreasing Z function does not hold true -- a real (inflation linked) curve. Because the actual size of the cash payments that will occur are unknown (as they are determined by the evolution of a price index, which is unknown) the arbitrage argument presented above does not hold. Thus, for a real curve the Z function is not necessarily decreasing (and empirically this phenomenon does on occasion occur).

f (t) = - d ln(Z(t))

(7)

dt

=

d r(t)t

(8)

dt

So f (t) = r(t) + r (t)t, so the forward rates will lie above the yield curve when the yield curve is normal, and below the yield curve when it is inverted. By integrating,1

t

r(t)t = f (s)ds

(9)

0

t

Z(t) = exp - f (s)ds

(10)

0

Also

riti - ri-1ti-1 =

1

ti

f (s)ds

(11)

ti - ti-1

ti - ti-1 ti-1

which shows that the average of the instantaneous forward rate over any of our intervals [ti-1, ti] is equal to the discrete forward rate for that interval. Finally,

t

r(t)t = ri-1ti-1 + f (s)ds, t [ti-1, ti]

(12)

ti-1

which is a crucial interpolation formula: given the forward function we easily find the risk free function.

1.1 Forward rates

If we can borrow at a known rate at time 0 to date t1, and we can borrow from t1 to t2 at a rate known and fixed at 0, then effectively we can borrow at a known rate at 0 until t2. Clearly

Z(0, t1)Z(0; t1, t2) = Z(0, t2)

(4)

is the no arbitrage equation: Z(0; t1, t2) is the forward discount factor for the period from t1 to t2--it has to be this value at time 0 with the information available at that time, to ensure no arbitrage.

The forward rate governing the period from t1 to t2, denoted f (0; t1, t2) satisfies

exp(-f (0; t1, t2)(t2 - t1)) = Z(0; t1, t2)

Immediately, we see that forward rates are positive (and this is equivalent to the discount function decreasing). We have either of

f

(0;

t1 ,

t2 )

=

-

ln(Z(0,

t2 )) t2

- -

ln(Z(0, t1

t1 ))

(5)

= r2t2 - r1t1 t2 - t1

(6)

Let the instantaneous forward rate for a tenor of t be denoted f(t), that is, f (t) = lim0 f (0; t, t + ), for whichever t this limit exists. Clearly then

2 Interpolation And Bootstrap Of Yield Curves--Not Two Separate Processes

As has been mentioned, many interpolation methods for curve construction are available. What needs to be stressed is that in the case of bootstrapping yield curves, the interpolation method is intimately connected to the bootstrap, as the bootstrap proceeds with incomplete information. This information is `completed' (in a non unique way) using the interpolation scheme.

In Hagan and West [2006] we illustrated this point using swap curves; here we will make the same points focusing on a bond curve. Suppose we have a reasonably small set of bonds that we want to use to bootstrap the yield curve. (To decide which bonds to include can be a non-trivial exercise. Excluding too many runs the risk of disposing of market information which is actually meaningful, on the other hand, including too many could result in a yield curve which is implausible, with a multitude of turning points, or even a bootstrap algorithm which fails to converge.) Recall that we insist that whatever instruments are included will be priced perfectly by the curve.

Typically some rates at the short end of the curve will be known. For example, some zero-coupon bonds might trade which give us exact rates. In some markets, where there is insufficient liquidity at the short end, some inter-bank money market rates will be used. Each bond and the curve must satisfy the following relationship:

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^

where

n

[A] = piZ(0; tsettle , ti)

i=0

? A is the all-in (dirty) price of the bond;

? tsettle is the date on which the cash is actually delivered for a pur-

chased bond;

? p0, p1, . . . , pn are the cash flows associated with a unit bond (typical-

ly

p0

=

e

c 2

,

pi

=

c 2

for

1

i

<

n

and

pn

=

1+

c 2

where

c is

the annual

coupon and e is the cum-ex switch);

? t0, t1, . . . , tn are the dates on which those cash flows occur.

On the left is the price of the bond trading in the market. On the

right is the price of the bond as stripped from the yield curve. We rewrite

this in the computationally more convenient form

n

[A ] Z(0, tsettle ) = piZ(0, ti)

(13)

i=0

Suppose for the moment that the risk free rates (and hence the dis-

count factors) have been determined at t0, t1, . . . , tn-1 . Then we solve Z(0, tn) easily as

Z(0, tn)

=

1 pn

n-1

[A ] Z(0, tsettle ) - piZ(0, ti)

i=0

which is written in the form of risk-free rates rather than discount fac-

tors as

rn

=

1 tn

ln pn - ln

n-1

[A ] e - -rsettle tsettle

pi e-ri ti

i=0

(14)

where the ti's are now denominated in years and the relevant day-count convention is being adhered to. Of course, in general, we do not know the

earlier rates, neither exactly (because it is unlikely that any money mar-

ket instruments expire exactly at ti) nor even after some interpolation (the rates for the smallest few ti might be available after interpolation, but the later ones not at all). However, as in the case of swap curves, (14)

suggests an iterative solution algorithm: we guess rn, indeed other expiry-date rates for other bonds, and take the rates already known from

e.g. the money market, and insert these rates into our interpolation algo-

rithm. We then determine rsettle and r0, r1, . . . , rn-1 . Next, we insert these rates into the right-hand side of (14) and solve for rn. We then take this new guess for this bond, and for all the other bonds, and again apply the

interpolation algorithm. We iterate this process. Even for fairly wild

curves (such as can often be the case in South Africa) this iteration will

reach a fixed point with accuracy of about 8 decimal places in 4 or 5 iter-

ations. This then is our yield curve.

3 How To Compare Yield Curve Interpolation Methodologies

In general, the interpolation problem is as follows: we have some data x as a function of time, so we have 1, 2, . . . , n and x1, x2, . . . , xn known. An interpolation method is one that constructs a continuous function x(t) satisfying x(i) = xi for i = 1, 2, . . . , n. In our setting, the x values

might be risk free rates, forward rates, or some transformation of these-- the log of rates, etc. Of course, many choices of interpolation function are possible -- according to the nature of the problem, one imposes requirements additional to continuity, such as differentiability, twice differentiability, conditions at the boundary, and so on.

The Lagrange polynomial is a polynomial of degree n - 1 which passes through all the points, and of course this function is smooth. However, it is well known that this function is inadequate as an interpolator, as it demonstrates remarkable oscillatory behaviour.

The typical approach is to require that in each interval the function is described by some low dimensional polynomial, so the requirements of continuity and differentiability reduce to linear equations in the coefficients, which are solved using standard linear algebraic techniques. The simplest example are where the polynomials are linear, and these methods are surveyed in ?4. However, these functions clearly will not be differentiable. Next, we try quadratics--however here we have a remarkable `zig-zag' instability which we will discuss. So we move on to cubics--or even quartics--they overcome these already-mentioned difficulties, and we will see these in ?5.

All of the interpolation methods considered in Hagan and West [2006] appear in the rows of Table 1.

We will restrict attention to the case where the number of inputs is reasonably small and so the bootstrapping algorithm is able to price the instruments exactly, and we restrict attention to those methods where the instruments are indeed always priced exactly.

The criteria to use in judging a curve construction and its interpolation method that we will consider are:

(a) In the case of yield curves, how good do the forward rates look? These are usually taken to be the 1m or 3m forward rates, but these are virtually the same as the instantaneous rates. We will want to have positivity and continuity of the forwards. It is required that forwards be positive to avoid arbitrage, while continuity is required as the pricing of interest sensitive instruments is sensitive to the stability of forward rates. As pointed out in McCulloch and Kochin [2000], `a discontinuous forward curve implies either implausible expectations about future short-term interest rates, or implausible expectations about holding period returns'. Thus, such an interpolation method should probably be avoided, especially when pricing derivatives whose value is dependent upon such forward values. Smoothness of the forward is desirable, but this should not be achieved at the expense of the other criteria mentioned here.

(b) How local is the interpolation method? If an input is changed, does the interpolation function only change nearby, with no or minor spill-over elsewhere, or can the changes elsewhere be material?

(c) Are the forwards not only continuous, but also stable? We can quantify the degree of stability by looking for the maximum basis point change in the forward curve given some basis point change (up or down) in one of the inputs. Many of the simpler methods can have this quantity determined exactly, for others we can only derive estimates.

(d) How local are hedges? Suppose we deal an interest rate derivative of a particular tenor. We assign a set of admissible hedging instruments, for example, in the case of a swap curve, we might (even should)

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decree that the admissible hedging instruments are exactly those instruments that were used to bootstrap the yield curve. Does most of the delta risk get assigned to the hedging instruments that have maturities close to the given tenors, or does a material amount leak into other regions of the curve?

We will now survey a handful of these methods, and highlight the issues that arise.

4 Linear Methods

4.1 Linear on rates

For ti-1 < t < ti the interpolation formula is

r(t)

=

t ti

- -

ti-1 ti-1

ri

+

ti

ti - t - ti-1

ri-1

(15)

Using (8) we get

f (t)

=

2t ti

- -

ti-1 ti-1

ri

+

ti - 2t ti - ti-1

ri-1

(16)

Of course f is undefined at the ti, as the function r(t)t is clearly not differentiable there. Moreover, in the actual rate interpolation formula, by the time t reaches ti, the import of rt-1 has been reduced to zero--that rate has `been forgotten'. But we clearly see that this is not the case for the forward, so the left and right limits f (ti+) and f (ti-) are different--the forward jumps. Furthermore, the choice of interpolation does not prevent negative forward rates: suppose we have the (t, r) points (1y, 8%) and (2y, 5%). Of course, this is a rather contrived economy: the one year interest rate is 8% and the one year forward rate in one year's time is 2%. Nevertheless, it is an arbitrage free economy. But using linear interpolation the instantaneous forwards are negative from about 1.84 years onwards.

4.2 Linear on the log of rates

Now for ti-1 t ti the interpolation formula is

ln(r(t))

=

t - ti-1 ti - ti-1

ln(ri) +

ti - t ti - ti-1

ln(ri-1 )

which as a rate formula is

t-ti-1

ti -t

r(t) = r r ti -ti-1 ti -ti-1

i

i-1

(17)

A simple objection to the above formula is that it does not allow negative interest rates. Also, the same argument as before shows that the forward jumps at each node, and similar experimentation will provide an example of a Z function which is not decreasing.

4.3 Linear on discount factors

Now for ti-1 t ti the interpolation formula is

Z(t)

=

t ti

- -

ti-1 ti-1

Zi

+

ti

ti - t - ti-1

Zi-1

which as a rate formula is

r(t) = -1 ln t - ti-1 e-riti + ti - t r-ri-1 ti-1

(18)

t

ti - ti-1

ti - ti-1

Again, the forward jumps at each node, and the Z function may not be decreasing.

4.4 Raw interpolation (linear on the log of discount factors)

This method corresponds to piecewise constant forward curves. This

method is very stable, is trivial to implement, and is usually the starting

point for developing models of the yield curve. One can often find mis-

takes in fancier methods by comparing the raw method with the more

sophisticated method.

By definition, raw interpolation is the method which has constant in-

stantaneous forward rates on every interval ti-1 < t < ti. From (11) we see

that that constant must be the discrete forward rate for the interval, so

f (t) =

ri ti -ri-1 ti-1 ti -ti-1

for ti-1

< t < ti. Then from (12) we have that

r(t)t

=

ri-1 ti-1

+

(t

-

ti-1 )

riti - ri-1ti-1 ti - ti-1

By writing the above expression with a common denominator of ti - ti-1, and simplifying, we get that the interpolation formula on that interval is

r(t)t

=

t ti

- -

ti-1 ti-1

ri ti

+

ti

ti - t - ti-1

ri-1

ti-1

(19)

which explains yet another choice of name for this method: `linear rt'; the method is linear interpolation on the points riti. Since ? riti is the logarithm of the capitalisation/discount factors, we see that calling this method `linear on the log of capitalisation factors' or `linear on the log of discount factors' is also merited.

This raw method is very attractive because with no effort whatsoever we have guaranteed that all instantaneous forwards are positive, because every instantaneous forward is equal to the discrete forward for the `parent' interval. As we have seen, this is an achievement not to be sneezed at. It is only at the points t1, t2, . . . , tn that the instantaneous forward is undefined, moreover, the function jumps at that point.

4.5 Piecewise linear forward

Having decided that the raw method is quite attractive, what happens if we try to remedy its only defect in the most obvious way? What we will do is that instead of the forwards being piecewise constant we will demand that they be a piecewise continuous linear function. What could be more natural than to simply ask to gently rotate the raw interpolants so that they are now not only piecewise linear, but continuous as well? Unfortunately, this very plausible requirement gives rise to at least two types of very unpleasant behaviour indeed. This is easily understood by means of an example.

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Curve, first scenario. 0.07

Forward, first scenario. 0.2

0.065

0.15

0.06

0.1

0.055

0.05

0.05

0

0.045 0

-0.05

5

10

0

5

10

0.065

Curve, second scenario.

Forward, second scenario. 0.2

0.06

0.15

0.1 0.055

0.05

0.05

0

0.045 0

-0.05

5

10

0

5

10

Figure 2: The piecewise linear forward method.

5.1 Quadratic splines

To complete a quadratic spline of a function x, we desire coefficients (ai, bi, ci) for 1 i n - 1. Given these coefficients, the function value at any term will be

x( ) = ai + bi( - i) + ci( - i)2 i i+1

(20)

The constraints will be that the interpolating function indeed meets the given data (and hence is continuous) and the entire function is differentiable. There are thus 3n - 4 constraints: n - 1 left hand function values to be satisfied, n - 1 right hand function values to be satisfied, and n - 2 internal knots where differentiability needs to be satisfied. However, there are 3n - 3 unknowns. With one degree of freedom remaining, it makes sense to require that the left-hand derivative at n be zero, so that the curve can be extrapolated with a horizontal asymptote.

Suppose we apply this method to the rates (so xi = ri). The forward curves that are produced are very similar to the piecewise linear forward curves--the curve can have a `zig-zag' appearance, and this zig-zag is subject to the same parity of input considerations as before.

So, next we try a cubic spline.

Firstly, suppose we have a curve with input zero coupon rates at every year node, with a value of r(t) = 5% for t = 1, 2, . . . , 5 and r(t) = 6% for t = 6, 7, . . . , 10. We must have f (t) = r(1) for t 1. In order to assure continuity, we see then we must have f (t) = r(i) for every i 5. Now, the discrete forward rate for [5, 6] is 11%. In order for the average of the piecewise linear function f on the interval [5, 6] to be 11% we must have that f (6) = 17%. And now in turn, the discrete forward rate for [6, 7] is 6% and so in order for the average of the piecewise linear function f on the interval [6, 7] to be 6% we must have that f (7) = -5%. This zig-zag feature continues recursively; see Figure 2. Note also the implausible shape of the actual yield curve itself.

Secondly, suppose now we include a new node, namely that r(6.5) = 6%. It is fairly intuitive that this imparts little new information2. Nevertheless, the bootstrapped curve changes dramatically. The `parity of the zig-zag' is reversed. So we see that the localness of the method is exceptionally poor.

5 Splines

The various linear methods are the simplest examples of polynomial splines: a polynomial spline is a function which is piecewise in each interval a polynomial, with the coefficients arranged to ensure at least that the spline coincides with the input data (and so is continuous). In the linear case that is all that one can do--the linear coefficients are now determined. If the polynomials are of higher degree, we can use up the degrees of freedom by demanding other properties, such as differentiability, twice differentiability, asymptotes at either end, etc.

The first thing we try is a quadratic spline.

5.2 Cubic splines

This time we desire coefficients (ai, bi, ci, di) for 1 i n - 1. Given these coefficients, the function value at any term will be

x( ) = ai + bi( - i) + ci( - i)2 + di( - i)3i i+1 (21)

As before we have 3n - 4 constraints, but this time there are 4n - 4 unknown coefficients. There are several possible ways to proceed to find another n constraints. Here are the ones that we have seen:

? xi = ri. The function is required to be twice differentiable, which for the same reason as previously adds another n - 2 constraints. For the final two constraints, the function is required to be linear at the extremes i.e. the second derivative of the interpolator at 1 and at n are zero. This is the so-called natural cubic spline.

? xi = ri. The function is again required to be twice differentiable; for the final two constraints we have that the function is linear on the left and horizontal on the right. This is the so-called financial cubic spline Adams [2001].

? xi = rii. The function is again required to be twice differentiable; for the final two constraints we have that this function is linear on the right and quadratic on the left. This is the quadratic-natural spline proposed in McCulloch and Kochin [2000].

? xi = ri. The values of bi for 1 < i < n are chosen to be the slope at i of the quadratic that passes through (j, rj) for j = i - 1, i, i + 1. The value of b1 is chosen to be the slope at 1 of the quadratic that passes through (j, rj) for j = 1, 2, 3; the value of bn is chosen likewise. This is the Bessel method [de Boor, 1978, 2001, Chapter IV], although often somewhat irregularly called the Hermite method by software vendors.

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? xi = rii. Again, Bessel interpolation.

? Going one step further, quartic splines. According to Adams [2001]

the quartic spline gives the smoothest interpolator of the forward

curve. The spline can proceed on instantaneous forward rates, this

time there are 5n - 5 unknowns and 3 additional conditions at 1 or

n required. Although one must ask: when does one actually have a

set of instantaneous forwards as inputs for interpolation?

Alternatively if we apply (9) then the inputs are risk free rates, and

the spline is of the form r( )

=

ai

+ bi

+ ci

+ di 2

+ ei 3

+ gi 4 , with

6n - 6 unknowns and 4 additional conditions required.

? xi = ri. The monotone preserving cubic spline of Hyman [1983]. The method specifies the values of bi for 1 i n, in a way to be dis-

cussed in more detail shortly.

Significant problems can become apparent when using some of these methods. The spline is supposed to alleviate the problem of oscillation seen when fitting a single polynomial to a data set (the Lagrange polynomial), nevertheless, significant oscillatory behaviour can still be present. Furthermore, the various types of clamping we see with some of the methods above (clamping refers to imposing conditions at the boundaries 1 or n) can compromise localness of the interpolator, sometimes grossly. In fact, the iterative procedure from ?2 often fails to converge for the quartic interpolation methods, and we exclude them from further analysis.

The method of Hyman is a method which attempts to address these problems. This method is quite different to the others; it is a local method--the interpolatory values are only determined by local behaviour, not global behaviour. This method ensures that in regions of monotonicity of the inputs (so, three successive increasing or decreasing values) the interpolating function preserves this property; similarly if the data has a minimum/maximum then the output interpolator will have a minimum/maximum at the node.

6 Monotone Convex

Many of the ideas of the method of Hyman will now have a natural development--the monotone convex method was developed to resolve the only remaining deficiency of Hyman [1983]. Very simply, none of the methods mentioned so far are aware that they are trying to solve a financial problem--indeed, the breeding ground for these methods is typically engineering or physics. As such, there is no mechanism which ensures that the forward rates generated by the method are positive, and some simple experimentation will uncover a set of inputs to a yield curve which give some negative forward rates under all of the methods mentioned here, as seen in Hagan and West [2006]. Thus, in introducing the monotone convex method, we use the ideas of Hyman [1983], but explicitly ensure that the continuous forward rates are positive (whenever the discrete forward rates are themselves positive).

The point of view taken in the monotone convex method is that the inputs are (or can be manipulated to be) discrete forwards belonging to intervals; the interpolation is not performed on the interest rate curve itself. We may have actual discrete forwards--FRA rates. On the other hand if we have interest rates r1, r2, . . . , rn for periods 1, 2, . . . , n then the

first thing we do is calculate fid

=

ri i -ri-1 i-1 i -i-1

for 1 i n, r0

= 0. (Here we

also check that these are all positive, and so conclude that the curve is

legal i.e. arbitrage free (except in those few cases where forward rates

may be negative). As an interpolation algorithm the monotone convex

method will now bootstrap a forward curve, and then if required recover

the continuum of risk free rates using (12).

One rather simple observation is that all of the spline methods we

saw in ?5 fail in forward extrapolation beyond the interval [1, n]. Clearly

if the interpolation is on rates then we will apply horizonal extrapolation

to the rate outside of that interval: r( ) = r1 for < 1 and r( ) = rn for

> n. So far so good. What happens to the forward rates? Perhaps sur-

prisingly we cannot apply the same extrapolation rule to the forwards, in

fact, we need to set f ( ) = r1 for < 1 and f ( ) = rn for > n--consider

(8). This makes it almost certain that the forward curve has a material

discontinuity at 1, and probably one at n too (the latter will be less se-

vere as the curve, either by design or by nature, probably has a horizontal

asymptote as n). In order to avoid this pathology, we now have terms 0 = 0, 1, . . . , n

and the generic interval for consideration is [i-1, i]. A `short rate' (in-

stantaneous) rate may be provided, if not, the algorithm will model one.

Usually the shortest rate that might be input will be an overnight rate, if

it is provided, the algorithm here simply has some `overkill'--there will

be an overnight rate and an instantaneous short rate--but it need not be

modified.

fid is the discrete rate which `belongs' to the entire interval [i-1, i]; it would be a mistake to model that rate as being the instantaneous rate at

i. Rather, we begin by assigning it to the midpoint of the interval, and

then modelling the instantaneous rate at i. as being on the straight line

that joins the adjacent midpoints. Let this rate f (i) be denoted fi. This explains (22). In (23) and (24) the values f0 = f (0) and fn = f (n) are selected so that f (0) = 0 = f (n). Thus

fi

=

i - i-1 i+1 - i-1

fid+1

+

i+1 - i i+1 - i-1

fid ,

for i = 1, 2, . . . , n - 1

(22)

f0

=

f1d

-

1 2

(f1

-

f1d )

(23)

fn

=

fnd

-

1 2

(fn-1

- fnd)

(24)

Note that if the discrete forward rates are positive then so are the fi for i = 1, 2, . . . , n - 1.

We now seek an interpolatory function f defined on [0, n] for f0, f1, . . . , fn that satisfies the conditions below (in some sense, they are arranged in decreasing order of necessity).

(i)

1 i -i-1

i i-1

f (t)dt

=

fid ,

so

the

discrete

forward

is

recovered

by

the

curve, as in (11).

(ii) f is positive.

(iii) f is continuous.

(iv) If fid-1 < fid < fid+1 then f ( ) is increasing on [i-1, i], and if fid-1 > fid > fid+1 then f ( ) is decreasing on [i-1, i].

Let us first normalise things, so we seek a function g defined on [0, 1] such that3

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g (0) 0 x = 0 = i?1

Figure 3: The function g.

x = 0 = i

g (1)

g(x) = f (i-1 + (i - i-1)x) - fid.

(25)

Before proceeding, let us give a sketch of how we will proceed. We will choose g to be piecewise quadratic in such a way that (i) is satisfied by construction. Of course, g is continuous, so (iii) is satisfied. As a quadratic, it is easy to perform an analysis of where the minimum or maximum occurs, and we thereby are able to apply some modifications to g to ensure that (iv) is satisfied, while ensuring (i) and (iii) are still satisfied.

Also, we see a posteriori that if the values of fi had satisfied certain constraints, then (ii) would have been satisfied. So, the algorithm will be to construct (22), (23) and (24), then modify the fi to satisfy those constraints, then construct the quadratics, and then modify those quadratics. Finally,

f ( ) = g

- i-1 i - i-1

+ fid.

(26)

Thus, the current choices of fi are provisional; we might make some

adjustments in order to guarantee the positivity of the interpolating

function f .

Here follow the details. We have only three pieces of information about

g: g(0) = fi-1 - fid, g(1) = fi - fid , and

1 0

g(x)dx

=

0.

We

postulate

a

func-

tional formg(x) =K +Lx + Mx2 , having 3 equations in 3 unknowns we get

100 K

g(0)

1 1 1 L = g(1) , and easily solve to find that

1

1 2

1 3

M

0

g(x) = g(0)[1 - 4x + 3x2] + g(1)[-2x + 3x2]

(27)

Note that by (22) that (iv) is equivalent to requiring that if fi-1 < fid < fi then f ( ) is increasing on [i-1, i], while if fi-1 > fid > fi then f ( ) is decreasing on [i-1, i]. This is equivalent to requiring that if g(0) and g(1) are of opposite sign then g is monotone. Now

g (x) = g(0)(-4 + 6x) + g(1)(-2 + 6x)

g (0) = -4g(0) - 2g(1)

g (1) = 2g(0) + 4g(1)

g being a quadratic it is now easy to determine, simply by inspecting g (0) and g (1), the behaviour of g on [0, 1]. The cases where g (0) = 0 and g (1) = 0 are crucial; these correspond to g(1) = -2g(0) and g(0) = -2g(1) respectively. These two lines divide the g(0)/g(1) plane into eight sectors. We seek to modify the definition of g on each sector, taking care that on the boundary of any two sectors, the formulae from those

g (1)

(ii)

(i)

(iv)

(iii)

D

g (0) (iii)

(iv)

(i)

(ii)

A

C

g(1) = -2g(0)

Figure 4: The reformulated possibilities for g.

g(0) = -2g(1) B

two sectors actually coincide (to preserve continuity). In actual fact the treatment for every diametrically opposite pair of sectors is the same, so we really have four cases to consider, as follows (refer Figure 4):

(i) In these sectors g(0) and g(1) are of opposite signs and g (0) and g (1) are of the same sign, so g is monotone, and does not need to be modified.

(ii) In these sectors g(0) and g(1) are also of opposite sign, but g (0) and g (1) are of opposite sign, so g is currently not monotone, but needs to be adjusted to be so. Furthermore, the formula for (i) and for (ii) need to agree on the boundary A to ensure continuity.

(iii) The situation here is the same as in the previous case. Now the formula for (i) and for (iii) need to agree on the boundary B to ensure continuity.

(iv) In these sectors g(0) and g(1) are of the same sign so at first it appears that g does not need to be modified. Unfortunately this is not the case: modification will be needed to ensure that the formula for (ii) and (iv) agree on C and (iii) and (iv) agree on D.

The origin is a special case: if g (0) = 0 = g (1) then g(x) = 0 for all x, and fid-1 = fid = fid+1 , and we put f ( ) = fid for [i-1, i]. So we proceed as follows:

(i) As already mentioned g does not need to be modified. Note that

on A we have g(x) = g(0)(1 - 3x2) and on B we have g(x) = g(0)

(1

-

3x

+

3 2

x2

).

(ii) A simple solution is to insert a flat segment, which changes to a

quadratic at exactly the right moment to ensure that

1 0

g(x)dx

=

0.

So we take

g(0)

for 0 x

g(x) =

g(0) + (g(1) - g(0))

x- 1-

2

for < x 1

(28)

=

1

+

g(0) 3 g(1) - g(0)

=

g(1) + 2g(0) g(1) - g(0)

(29)

Note that - 0 as g(1) - -2g(0), so the interpolation formula reduces to g(x) = g(0)(1 - 3x2) at A, as required.

76

WILMOTT magazine

TECHNICAL ARTICLE 2

(iii) Here again we insert a flat segment. So we take

g(x) =

g(1) + (g(0) - g(1))

-x

2

for 0 < x <

(30)

g(1)

for x < 1

=

3

g(1) g(1) - g(0)

(31)

Note

that

-

1

as

g(1)

-

-

1 2

g(0),

so

the

interpolation

formula

reduces

to

g(x)

=

g(0)(1

-

3x

+

3 2

x2

)

at

B,

as

required.

(iv) We want a formula that reduces in form to that defined in (ii) as we

approach C, and to that defined in (iii) as we approach D. This sug-

gests

A + (g(0) - A) g(x) = A + (g(1) - A)

-x

x- 1-

2 2

for for

0 -fid since fi-1, fi

are positive. Thus the inequality is satisfied at the endpoints of the inter-

val. Now, in regions (i), (ii) and (iii), g is monotone, so those regions are

fine.

In region (iv) g is not monotone. g is positive at the endpoints and has

a minimum of A (as in (34)) at the x-value (as in (33)). So, it now suffices

to

prove that

g(0)g(1) g(0)+g(1)

<

fid . This is the case if fi-1, fi

<

3fid .

To

see

this,

note

that then 0 < g(0), g(1) < 2fid and the result follows, since if 0 < y, z < 2a

then

y+z yz

=

1 z

+

1 y

>

1 2a

+

1 2a

=

1 a

and so

yz y+z

> a.

We choose the slightly stricter condition fi-1, fi < 2fid . Thus, our algo-

rithm is

(1) Determine the fid from the input data. (2) Define fi for i = 0, 1, . . ., n as in (22), (23) and (24). (3) If f is required to be everywhere positive, then collar f0 between 0

and 2f1d, for i = 1, 2, . . . , n - 1 collar fi between 0 and 2min(fid, fid+1), and collar fn between 0 and 2fnd. If f is not required to be everywhere positive, simply omit this step. (4) Construct g with regard to which of the four sectors we are in. (5) Define f as in (26). (6) If required recover r as in (12). Integration formulae are easily established as the functions forms of g are straightforward.

Pseudo-code for this recipe is provided in an Appendix. Working code for this interpolation scheme is available from the second author's website.

6.2 Amelioration

In Hagan and West [2006] an enhancement of this method is considered where the curve is ameliorated (smoothed). This is achieved by making the interpolation method slightly less local i.e. by using as inputs not only neighbouring information but also information which is two nodes away.

7 Hedging

We can now ask the question: how do we use the instruments which have been used in our bootstrap to hedge other instruments? In general

1

1

1

1

0

0

0

0

-1

-1

-1

-1

0

0.5

10

0.5

10

0.5

10

0.5

1

1

1

1

1

0

0

0

0

-1

-1

-1

-1

0

0.5

10

0.5

10

0.5

10

0.5

1

1

1

1

1

0

0

0

0

-1

-1

-1

-1

0

0.5

10

0.5

10

0.5

10

0.5

1

1

1

1

1

0

0

0

0

-1

-1

-1

-1

0

0.5

10

0.5

10

0.5

10

0.5

1

1

1

1

1

0

0

0

0

-1

-1

-1

-1

0

0.5

10

0.5

10

0.5

10

0.5

1

Figure 5: The g function as we cross the boundaries. From left to right: boundaries A, B, C and D. From top to bottom: approaching the boundary, at the boundary (central), leaving the boundary. Only at the boundary of Cand D are there discontinuities.

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